Given that the degree of splitting field of a polynomial $f(x)$ over $\mathbb{Q}[x]$ is equal to $n!$ where $n$ is the degree of $f$, $n>2$. If $\alpha$ is a root of $f$ in the splitting field, then we have to show that $\mathbb{Q}(\alpha^4)=\mathbb{Q}(\alpha)$.
My attempt: $\mathbb{Q}(\alpha^4)\subseteq\mathbb{Q}(\alpha)$ is obvious.Also, $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^4)]=4$ as $x^4-\alpha^4$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}(\alpha^4)$. So, $n$ must be a multiple of $4$, by Tower Law. I have come to a dead end it seems Please guide me.
I will simply try to join the dots mentioned in the comments.
First let us see why the conjugates of $\alpha^4$ are exactly $\alpha_i^4$, $ i=1,2,\cdots,n$.If $g(x)$ is the minimal polynomial of $\alpha^4$, then $\alpha$ is the root of $g(x^4)$ but $\alpha$ has $f(x)$ to be the minimal polynomial over $\mathbb{Q}$ because if $f(x)$ is reducible then the splitting field will have a degree strictly less than $n!$. Thus, $f(x)$ must divide $g(x^4)$, implying all the roots of $f(x)$, namely the $\alpha_i$'s must also be roots of $g(x^4)$. That is $\alpha_i^4$ is a root of $g(x)$. Hence the claim.
Now, all we have to show is that these conjugates are distinct.
$\alpha_j^4=\alpha_k^4,j\neq k$ implies four choices $\alpha_j=i^t\alpha_k$ , $t=1,2,3,4$ of which we claim none is possible. Let us first remove the obvious $t=4$ in which case $\alpha_j=\alpha_k$, not true.
For t=1, the minimal polynomial of $\alpha_j$ over $\mathbb{Q}(\alpha_k)$ is $x^2+\alpha_k^2$.Therefore, $[\mathbb{Q}(\alpha_j,\alpha_k):\mathbb{Q}(\alpha_k)]=2$ contradicting $[\Bbb{Q}(\alpha_j,\alpha_k):\Bbb{Q}]<n(n-1)$ , since $n>3$, so $n(n-1)>2n=[\Bbb{Q}(\alpha_j,\alpha_k):\Bbb{Q}]$.
And similarly the other cases.